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I have $X_1,X_2 ... X_n$ Gamma-distributed r.v with density:

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First of all I showed, that Gamma distribution belongs to exponential family and can be represented in form

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I found, that for Gamma distribution $B(\theta) = log(\theta)$ where $\theta$ is $\alpha/\beta$

Then we know, that moment generating function of the random variable $T(X)$ is given by $$M(t)=\exp\{B(\theta+t)-B(\theta)\}$$

Natural sufficient statistics for Gamma distribution is:

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But I didn't understand what should I do next? Because Natural sufficient statistics is two dimensional it confuses me:

$M(T) = exp[log(\theta + T) - log(\theta)]$

and $T = [s,t]^\intercal$

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    $\begingroup$ Your representation of an exponential distribution is (a) unidimensional and (b) based on $y$ being the sufficient statistic. Check the literature or even the Wikipedia page for proper definitions. $\endgroup$ – Xi'an Dec 6 '17 at 18:15
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As already recalled in the answer to a previous question of yours and well-explained on Wikipedia, for any exponential family, there exists a parameterisation such that the density of the family is$$f(x|\theta)=\exp\{\theta\cdot T(x)-\Psi(\theta)\}$$wrt a constant measure $\text{d}\mu(x)$, where the components of $T(\cdot)$ are linearly independent. The $\cdot$ in $\theta\cdot T(x)$ represents the scalar product, which means both $\theta$ and $T(x)$ are vector of the same dimension.

When you consider the density of the $\text{Ga}(\alpha,\beta)$ distribution,$$f(x|\alpha,\beta)=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}\exp\{-\beta x\}\mathbb{I}_{(0,\infty)}(x)$$moving terms into the exponential leads to \begin{align*} f(x|\alpha,\beta)&=\dfrac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}\exp\{-\beta x\}\mathbb{I}_{(0,\infty)}(x)\\ &=\exp\{\log[\beta^\alpha] - \log \Gamma(\alpha)+[\alpha-1]\log x-\beta x\}\mathbb{I}_{(0,\infty)}(x)\\ &=\exp\{\underbrace{[\alpha-1]\log x-\beta x}_{\theta\cdot T(x)} + \underbrace{\log[\beta^\alpha] - \log \Gamma(\alpha)}_{\Psi(\theta)}\}\mathbb{I}_{(0,\infty)}(x)\\ \end{align*} which suggests $\theta=(\alpha-1,\beta)$ as natural parameter and $T(x)=(\log x,-x)$ as sufficient statistic. (But other choices are possible.)

Hence, for this parameterisation of the $\text{Ga}(\alpha,\beta)$ distribution as an exponential family, \begin{align*} \Psi(\theta)&=-\log[\beta^\alpha] + \log \Gamma(\alpha)\\ &=-(\alpha-1+1)\log \beta + \log \Gamma(\alpha-1+1)\\ &= -[1+\theta_1]\log \theta_2 +\log \Gamma(1+\theta_1) \end{align*} as the cumulant moment function.

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